6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

A person sold two cows each for Rs.9900. If he gained 10% on one and lost 20% on the other, then which of the following is true?

A) He Gained Rs. 200	B) He lost Rs. 200
C) He neither gained nor lost	D) None of the above

Answer & Explanation Answer: D) None of the above

Explanation:

The CP of profitable cow = 9900/1.1 = 9000

and profit = Rs. 900

The CP of loss yielding cow = 9900/0.8 = 12375

and loss = Rs. 2475

so, the net loss = 2475 - 900 = 1575

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Filed Under: Profit and Loss
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

26 19404

Q:

A tailor has 37.5 metres of cloth and he has to make 8 pieces out of a metre of cloth. How many pieces can he make out this cloth ?

A) 300	B) 360
C) 400	D) 450

Answer & Explanation Answer: A) 300

Explanation:

Length of each piece = (1/8) m = 0.125 m

Required no of pieces = (37.5/0.125) = (375 * 100) / 125 = 300.

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Filed Under: Decimal Fractions

36 19362

Q:

For any two numbers m , n ; (m+n) : (m-n) : mn = 7: 1: 60 . Find the value of 1/m : 1/n

A) 4:3	B) 8:7
C) 3:4	D) 7:8

Answer & Explanation Answer: C) 3:4

Explanation:

$\frac{m + n}{m - n} = \frac{7 x}{x} \Rightarrow \frac{m}{n} = \frac{4 x}{3 x}$

Again $m n = 12 x^{2}$

and mn =60x

so, $60 x = 12 x^{2} \Rightarrow x = 5$

=> m= 20 and n= 15

Hence, $\frac{1}{m} : \frac{1}{n} = \frac{1}{20} : \frac{1}{15} = 3 : 4$

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Filed Under: Ratios and Proportions

47 19359

Q:

∆PQR is right angled at Q. If m∠P = 60 deg, then find the value of (secR + 1/2).

A) (2√2+√3)/2	B) (√3+4)/2√3
C) √3+2	D) 5/6

Answer & Explanation Answer: B) (√3+4)/2√3

Explanation:

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Filed Under: Volume and Surface Area
Exam Prep: Bank Exams

0 19322

Q:

Eight years ago, Ajay's age was 4/3 times that of Vijay. Eight years hence, Ajay's age will be 6/5 times that of Vijay. What is the present age of Ajay ?

A) 41 yrs	B) 40 yrs
C) 37 yrs	D) 33 yrs

Answer & Explanation Answer: B) 40 yrs

Explanation:

Let the present ages of Ajay and Vijay be 'A' and 'V' years respectively.

=> A - 8 = 4/3 (V - 8) and A + 8 = 6/5 (V + 8)

=> 3/4(A - 8) = V - 8 and 5/6(A + 8) = V + 8

V = 3/4 (A - 8) + 8 = 5/6 (A + 8) - 8

=> 3/4 A - 6 + 8 = 5/6 A + 20/3 - 8

=> 10 - 20/3 = 10/12 A - 9/12 A

=> 10/3 = A/12 => A = 40.

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Filed Under: Problems on Ages
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

26 19272

Q:

Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is :

A) 12	B) 137
C) 14	D) 15

Answer & Explanation Answer: D) 15

Explanation:

Let the three integers be x, x + 2 and x + 4. Then, 3x = 2 (x + 4) + 3 <=> x = 11.

Third integer = x + 4 = 15.

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Filed Under: Problems on Numbers

38 19188

Q:

3 x 0.3 x 0.03 x 0.003 x 30 =?

A) 0.0000243	B) 0.000243
C) 0.00243	D) 0.0243

Answer & Explanation Answer: C) 0.00243

Explanation:

3 x 3 x 3 x 3 x 30 = 2430. Sum of decimal places = 6

Therefore, 3 x 0.3 x 0.03 x 0.003 x 30 = 0.002430 = 0.00243

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Filed Under: Decimal Fractions

72 19179

Q:

A child has three different kinds of chocolates costing Rs.2, Rs.5, Rs.10. He spends total Rs. 120 on the chocolates. what is the minimum possible number of chocolates he can buy, if there must be atleast one chocolate of each kind?

A) 22	B) 19
C) 17	D) 15

Answer & Explanation Answer: C) 17

Explanation:

Minimum number of chocolates are possible when he purchases maximum number of costliest chocolates.

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